Last name: name: 1 Quiz 9 (Notes, books, and calculators are not authorized) Show all your work in the blank space you are given on the exam sheet. Answers with no justification will not be graded. Question 1: (a) Find the derivative of f (z) = 5z 3 +3iz−3+i and specify where f is holomorphic f 0 (z) = 15z 2 + 3i. f is holomorphic in C. (b) Find the derivative of F (z) = 2z+i z+2i and specify where F is holomorphic. 2(z + 2i) − (2z + i) 3i = . 2 (z + 2i) (z + 2i)2 F 0 (z) = F is holomorphic in C \ {− 21 i}. Question 2: Let f (z) = z 3 . Compute integral arc of radius 1 starting at 1 and ending at −1. R C f (z)dz where C is the upper semi-circular solution 1: Owing to a corollary of Cauchy’s Theorem the integral is independent of the path that is chosen to connect 1 to −1. We may choose the line γ(t) = 1 − 2t, t ∈ [0, 1], then Z Z f (z)dz = C 1 −2t3 dt = 0 0 solution 2: Another possibility consists of realizing that f (z) = g 0 (z) where g(z) = 41 z 4 , then Z Z f (z)dz = g 0 (z)dz = g(−1) − g(1) = 0 C C solution 3: Compute the integral using the definition of the semi-circular arc: γ(t) = eiπt , with t ∈ [0, 1]. Z Z f (z)dz = C 0 1 e3iπt iπeiπt dt == 1 4 Z 0 1 4iπe4iπt dt = 1 4iπ (e − 1) = 0. 4 2 Quiz 9, November 29, 2012 Question 3: Compute −5 + 2i. R dz C z−2 where C is the perimeter of the square of width 3 centered at 1 is holomorphic in the open set C \ {2} and the square of width 3 centered The function h(z) = z−2 at −5 + 2i is in C \ {2}. Then the integral is zero owing to Cauchy’s Theorem. R √ √ √ θ Question 4: Let us define z = rei 2 for θ ∈ [−π, π], where z = reiθ . Compute C zdz where C is the straight line connecting 1 − i to 1 + i. √ Solution 1: The definition of z implies that this function is holomorphic in the complex plane minus the axis (−∞, 0). Then Z Z √ √ zdz = zdz, C √ L where L is the portion of circular arc of radius 2 centered√at 0 and joining 1 − i to 1 + i. A π π possible parameterization of L is defined as follows: γ(t) = 2ei(t 2 − 4 ) , with t ∈ [0, 1]. Observe π π that t 2 − 4 ∈ [−π, π] when t spans [0, 1]. Then Z C √ Z 1 π π π 2 3 3π 2 i ei(t 2 − 4 ) dt = 2 4 i zdz = 2 e 2 3 4 0 3π 2 3 −i 3π i 3π 2 3 i 3π = 2 4 e 8 (e 4 − 1) = 2 4 (e 8 − e−i 8 ). 3 3 1 4 π i(t π 4−8) 1 2 Z 1 ei(t 3π 3π 4 − 8 ) 0 In conclusion Z C √ 4 3 3π zdz = i 2 4 sin( ) 3 8 3 √ Solution 2: Observe that z = 23 dzdz2 . Hence Z √ 3 3 3π 2 2 3 3π zdz = ((1 + i) 2 − (1 − i) 2 ) = 2 4 (ei 8 − e−i 8 ). 3 3 C dt